Calculating Translational Energy in a Nitrogen Gas Vessel

Click For Summary
SUMMARY

The discussion focuses on calculating the total translational kinetic energy of nitrogen gas in a 5L vessel at 27 degrees Celsius and 3 atm pressure. The total translational energy is derived using the equation Etrans = (3/2)NkBT, resulting in a value of 1669.9 J/K. The average kinetic energy per molecule is calculated using the simplified equation (3/2)kBT, which provides a clear distinction between total and average energy calculations.

PREREQUISITES
  • Understanding of the Ideal Gas Law and its applications
  • Familiarity with the concepts of translational kinetic energy
  • Knowledge of Avogadro's number and molar mass calculations
  • Basic understanding of Boltzmann's constant and its significance in thermodynamics
NEXT STEPS
  • Learn how to apply the Ideal Gas Law to different gas conditions
  • Study the relationship between temperature and kinetic energy in gases
  • Explore the concept of molecular mass and its calculation for different gases
  • Investigate the implications of translational energy in real-world applications, such as thermodynamics
USEFUL FOR

Students studying thermodynamics, physicists, chemists, and anyone interested in understanding gas behavior and energy calculations in molecular systems.

trah22
Messages
44
Reaction score
0

Homework Statement


a 5L verssel contains notrogen gas at 27 degrees celsius and 3 atm. Find a) total translational kinetic energy of gas molecules and b)average kinetic energy per molecule


Homework Equations


Etrans=3/2nRT and etrans=3/2NkT, N=6.022x10^23 n=mass/molarmass,
R=.0821

The Attempt at a Solution



for part a)3/2(not sure how to get n)(3(.0821)(300)

i multiplied .0821 by 3 because the pressure in the question is 3 atm and one atm is .0821 not sure if this is right... i don't quite understand what to do with the volume for part b, i know i have to use avogadro's number since its asking per molecule but I am not sure how to apply it...could someone help me out
 
Physics news on Phys.org
trah22 said:
2. Homework Equations [/b]
Etrans=3/2nRT and etrans=3/2NkT, N=6.022x10^23 n=mass/molarmass,
R=.0821

In your equation N is *not* Avagadro's number. N is the number of Nitrogen molecules. You need to look up the density of Nitrogen gas and multiply that by the given volume to find the total mass of the gas. Then lookup the mass of a single molecule of Nitrogen and divide the total mass of the gas by the mass of a single molecule to find N.

Once you have N, multiply it by 3/2 then multiply it by "k" (Boltzmann's Constant) then multiply it by the temperature in Kelvin. This will give you the total translational energy of the 5 liters of Nitrogen. Cheers
 
the density is 1.251 g/l so multiplying that by 5 gives 6.255 but i don't understand one thing, after u divide this mass by 14(molarmass of Nitrogen) don't u get the number of moles because n=m/M so wouldn't u then multiply that by avogadros number to get the total number of molecules?

could u also tell me the difference between find the total trslational kinetic energy of the gas molecules vs the average KE per molecule, i understand that Etrans=nKt gives total translational but i don't know what to apply to calculate the average KE per molecule
 
Last edited:
trah22 said:
the density is 1.251 g/l so multiplying that by 5 gives 6.255 but i don't understand one thing, after u divide this mass by 14(molarmass of Nitrogen) don't u get the number of moles because n=m/M so wouldn't u then multiply that by avogadros number to get the total number of molecules?

The mass of one mole of Nitrogen is 14 grams. That is not the same as the mass of one molecule of Nitrogen. The mass of one molecule of Nitrogen is 14/(6.02*10^23) grams.

So, yes, if you dived 6.255 grams by 14 grams you do get the number of *moles* and then multiplying by Avogadro's number gives the number of molecules.

That works, but that's not what I told you to do. If you do what I told you to do, which was to divide 6.255 grams by the mass of one nitrogen *molecule*, you will also get the same answer since
<br /> N=\frac{6.255}{m_{\textrm{mole}}}N_{\textrm{avogadro}}=\frac{6.255}{14}N_{\textrm{avogadro}}=<br /> \frac{6.255}{14/N_{\textrm{avogadro}}}=\frac{6.255}{m_{\textrm{molecule}}}<br />

Regardless how you choose to break up the calculation. The next step is to take the above value of N and multiply it by 3kT/2. Cheers.
 
alright thanks for clearing that up,

kb=1.38x10^-23
N=2.689x10^23

Etrans=3NkT/2
=3(2.689x10^23)(1.38x10^-23)(300K)/2
=1669.9J/K

this is the amount of total translational energy of the gas molecules, for part b of the question (finding the average kinetic energy per molecule) is correct equation (im looking at my notes) just 3kbT/2 without any N, but that doesn't quite make sense...
 
trah22 said:
alright thanks for clearing that up,

kb=1.38x10^-23
N=2.689x10^23

Etrans=3NkT/2
=3(2.689x10^23)(1.38x10^-23)(300K)/2
=1669.9J/K

this is the amount of total translational energy of the gas molecules, for part b of the question (finding the average kinetic energy per molecule) is correct equation (im looking at my notes) just 3kbT/2 without any N, but that doesn't quite make sense...

yeah, it does make sense. The total translational energy is
<br /> \frac{3}{2}Nk_bT<br />
which you just found. This is the *total* translational energy. I.e. the sum of the translational kinetic energies of all of the molecules.

The "average kinetic energy per molecule" is defined to mean "that quantity which, when multiplied by the total number of molecules, gives the total energy." It's just the total energy \frac{3}{2}Nk_bT divided by the total number of molecules N. In other words:
<br /> \left(\frac{3}{2}Nk_bT\right)/N<br />

But that *is* just \frac{3}{2}k_bT
 
thanks for that great explanation man, u cleared it up completely for me:cool:
 
I'm glad that I helped. Cheers.
 

Similar threads

Replies
5
Views
935
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 39 ·
2
Replies
39
Views
4K
Replies
7
Views
2K
Replies
3
Views
14K
Replies
1
Views
6K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K